Let F(adsbygoogle = window.adsbygoogle || []).push({}); _{n}= 2^{2n}+ 1 be the nth Fermat number and suppose that p|F_{n}, where p is a prime (possibly F_{n}itself). Show that [itex] 2^{2^{n+1}} \equiv 1 (mod p) [/itex] so that [itex] ord_p(2)|2^{n+1} [/itex]. Use this to show that [itex] ord_p(2) = 2^{n+1} [/itex].

So far i have shown that [itex] 2^{2^{n+1}} \equiv 1 (mod p) [/itex] so that [itex] ord_p(2)|2^{n+1} [/itex]. But what I'm having trouble showing now is that 2^{n+1}is the smallest possible number k such that [itex] 2^k \equiv 1 (mod p) [/itex]. I know that the order of 2 must be some power of 2 so I have tried to use a contradiction argument assuming that there exists a 2^{k}such that [itex] 2^{2^k} \equiv 1 (mod p) [/itex]. then i have [itex] 2^{2^{n+1}} \equiv 2^{2^k} (mod p) [/itex] so [itex] 2^{2^{n+1}-2^k} \equiv 1 (mod p) [/itex]. but i can't seem to find a contradiction.

can someone give me some hints to continue? thanks.

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# Problem about Fermat numbers and order of an element

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